创建与数字范围匹配的正则表达式的函数

Question

我正在使用 Amazon Mechanical Turk API，它只允许我使用正则表达式来过滤数据字段。

我想向函数输入一个整数范围，例如 256-311 或 45-1233，并返回仅匹配该范围的正则表达式。

匹配 256-321 的正则表达式将是：

\b((25[6-9])|(2[6-9][0-9])|(3[0-1][0-9])|(32[0-1]))\b

这部分相当简单，但我在创建此正则表达式的循环时遇到问题。

我正在尝试构建一个如下定义的函数：

function getRangeRegex( int fromInt, int toInt)
{

      return regexString;
}

我浏览了整个网络，令我惊讶的是，过去似乎没有人解决过这个问题。这是一个难题...

感谢您的宝贵时间。

Answer 1

这是一个快速破解：

<?php

function regex_range($from, $to) {

  if($from < 0 || $to < 0) {
    throw new Exception("Negative values not supported"); 
  }

  if($from > $to) {
    throw new Exception("Invalid range $from..$to, from > to"); 
  }

  $ranges = array($from);
  $increment = 1;
  $next = $from;
  $higher = true;

  while(true) {

    $next += $increment;

    if($next + $increment > $to) {
      if($next <= $to) {
        $ranges[] = $next;
      }
      $increment /= 10;
      $higher = false;
    }
    else if($next % ($increment*10) === 0) {
      $ranges[] = $next;
      $increment = $higher ? $increment*10 : $increment/10;
    }

    if(!$higher && $increment < 10) {
      break;
    }
  }

  $ranges[] = $to + 1;

  $regex = '/^(?:';

  for($i = 0; $i < sizeof($ranges) - 1; $i++) {
    $str_from = (string)($ranges[$i]);
    $str_to = (string)($ranges[$i + 1] - 1);

    for($j = 0; $j < strlen($str_from); $j++) {
      if($str_from[$j] == $str_to[$j]) {
        $regex .= $str_from[$j];
      }
      else {
        $regex .= "[" . $str_from[$j] . "-" . $str_to[$j] . "]";
      }
    }
    $regex .= "|";
  }

  return substr($regex, 0, strlen($regex)-1) . ')$/';
}

function test($from, $to) {
  try {
    printf("%-10s %s\n", $from . '-' . $to, regex_range($from, $to));
  } catch (Exception $e) {
    echo $e->getMessage() . "\n";
  }
}

test(2, 8);
test(5, 35);
test(5, 100);
test(12, 1234);
test(123, 123);
test(256, 321);
test(256, 257);
test(180, 195);
test(2,1);
test(-2,4);

?>

它会产生：

2-8        /^(?:[2-7]|8)$/
5-35       /^(?:[5-9]|[1-2][0-9]|3[0-5])$/
5-100      /^(?:[5-9]|[1-9][0-9]|100)$/
12-1234    /^(?:1[2-9]|[2-9][0-9]|[1-9][0-9][0-9]|1[0-2][0-3][0-4])$/
123-123    /^(?:123)$/
256-321    /^(?:25[6-9]|2[6-9][0-9]|3[0-2][0-1])$/
256-257    /^(?:256|257)$/
180-195    /^(?:18[0-9]|19[0-5])$/
Invalid range 2..1, from > to
Negative values not supported

未经正确测试，使用风险自负！

是的，在许多情况下生成的正则表达式可以写得更紧凑，但我将其作为读者的练习:)

Answer 2

对于像我一样正在寻找上面伟大的 @Bart Kiers 作品的 javascript 版本的人

//Credit: Bart Kiers 2011
function regex_range(from, to){
        if(from < 0 || to < 0) {
            //throw new Exception("Negative values not supported"); 
            return null;
        }
        if(from > to) {
            //throw new Exception("Invalid range from..to, from > to"); 
            return null;
        }

        var ranges = [];
        ranges.push(from);
        var increment = 1;
        var next = from;
        var higher = true;

        while(true){
            next += increment;
            if(next + increment > to) {
                if(next <= to) {
                    ranges.push(next);
                }
                increment /= 10;
                higher = false;
            }else{ 
                if(next % (increment*10) == 0) {
                    ranges.push(next);
                    increment = higher ? increment*10 : increment/10;
                }
            }

            if(!higher && increment < 10) {
                break;
            }
        }

        ranges.push(to + 1);
        var regex = '/^(?:';

        for(var i = 0; i < ranges.length - 1; i++) {
            var str_from = ranges[i];
            str_from = str_from.toString();
            var str_to = ranges[i + 1] - 1;
            str_to = str_to.toString();
            for(var j = 0; j < str_from.length; j++) {
                if(str_from[j] == str_to[j]) {
                    regex += str_from[j];
                }
                else {
                    regex += "[" + str_from[j] + "-" + str_to[j] + "]";
                }
            }
            regex += "|";
        }

        return regex.substr(0, regex.length - 1 ) + ')$/';
    }

Answer 3

的 PHP 端口

class RegexRangeNumberGenerator {

    static function parse($min, $max, $MatchWholeWord = FALSE, $MatchWholeLine = FALSE, $MatchLeadingZero = FALSE) {
        if (!is_int($min) || !is_int($max) || $min > $max || $min < 0 || $max < 0) {
            return FALSE;
        }
        if ($min == $max) {
            return self::parseIntoPattern($min, $MatchWholeWord, $MatchWholeLine, $MatchLeadingZero);
        }
        $s = [];
        $x = self::parseStartRange($min, $max);
        foreach ($x as $o) {
            $s[] = self::parseEndRange($o[0], $o[1]);
        }
        $n = self::reformatArray($s);
        $h = self::parseIntoRegex($n);
        return self::parseIntoPattern($h, $MatchWholeWord, $MatchWholeLine, $MatchLeadingZero);
    }

    static private function parseIntoPattern($t, $MatchWholeWord = FALSE, $MatchWholeLine = FALSE, $MatchLeadingZero = FALSE) {
        $r = ((is_array($t)) ? implode("|", $t) : $t);
        return (($MatchWholeLine && $MatchLeadingZero) ? "^0*(" . $r . ")$" : (($MatchLeadingZero) ? "0*(" . $r . ")" : (($MatchWholeLine) ? "^(" . $r . ")$" : (($MatchWholeWord) ? "\\b(" . $r . ")\\b" : "(" . $r . ")"))));
    }

    static private function parseIntoRegex($t) {
        if (!is_array($t)) {
            throw new Exception("Argument needs to be an array!");
        }
        $r = [];
        for ($i = 0; $i < count($t); $i++) {
            $e = str_split($t[$i][0]);
            $n = str_split($t[$i][1]);
            $s = "";
            $o = 0;
            $h = "";
            for ($a = 0; $a < count($e); $a++) {
                if ($e[$a] === $n[$a]) {
                    $h .= $e[$a];
                } else {
                    if ((intval($e[$a]) + 1) === intval($n[$a])) {
                        $h .= "[" . $e[$a] . $n[$a] . "]";
                    } else {
                        if ($s === ($e[$a] . $n[$a])) {
                            $o++;
                        }
                        $s = $e[$a] . $n[$a];
                        if ($a == (count($e) - 1)) {
                            $h .= (($o > 0) ? "{" . ($o + 1) . "}" : "[" . $e[$a] . "-" . $n[$a] . "]");
                        } else {
                            if ($o === 0) {
                                $h .= "[" . $e[$a] . "-" . $n[$a] . "]";
                            }
                        }
                    }
                }
            }
            $r[] = $h;
        }
        return $r;
    }

    static private function reformatArray($t) {
        $arrReturn = [];
        for ($i = 0; $i < count($t); $i++) {
            $page = count($t[$i]) / 2;
            for ($a = 0; $a < $page; $a++) {
                $arrReturn[] = array_slice($t[$i], (2 * $a), 2);
            }
        }
        return $arrReturn;
    }

    static private function parseStartRange($t, $r) {
        if (strlen($t) === strlen($r)) {
            return [[$t, $r]];
        }
        $break = pow(10, strlen($t)) - 1;
        return array_merge([[$t, $break]], self::parseStartRange($break + 1, $r));
    }

    static private function parseEndRange($t, $r) {
        if (strlen($t) == 1) {
            return [$t, $r];
        }
        if (str_repeat("0", strlen($t)) === "0" . substr($t, 1)) {
            if (str_repeat("0", strlen($r)) == "9" . substr($r, 1)) {
                return [$t, $r];
            }
            if ((int) substr($t, 0, 1) < (int) substr($r, 0, 1)) {
                $e = intval(substr($r, 0, 1) . str_repeat("0", strlen($r) - 1)) - 1;
                return array_merge([$t, self::strBreakPoint($e)], self::parseEndRange(self::strBreakPoint($e + 1), $r));
            }
        }
        if (str_repeat("9", strlen($r)) === "9" . substr($r, 1) && (int) substr($t, 0, 1) < (int) substr($r, 0, 1)) {
            $e = intval(intval((int) substr($t, 0, 1) + 1) . "" . str_repeat("0", strlen($r) - 1)) - 1;
            return array_merge(self::parseEndRange($t, self::strBreakPoint($e)), [self::strBreakPoint($e + 1), $r]);
        }
        if ((int) substr($t, 0, 1) < (int) substr($r, 0, 1)) {
            $e = intval(intval((int) substr($t, 0, 1) + 1) . "" . str_repeat("0", strlen($r) - 1)) - 1;
            return array_merge(self::parseEndRange($t, self::strBreakPoint($e)), self::parseEndRange(self::strBreakPoint($e + 1), $r));
        }
        $a = (int) substr($t, 0, 1);
        $o = self::parseEndRange(substr($t, 1), substr($r, 1));
        $h = [];
        for ($u = 0; $u < count($o); $u++) {
            $h[] = ($a . $o[$u]);
        }
        return $h;
    }

    static private function strBreakPoint($t) {
        return str_pad($t, strlen(($t + 1)), "0", STR_PAD_LEFT);
    }
}

RegexNumericRangeGenerator测试结果

2-8         ^([2-8])$
5-35        ^([5-9]|[12][0-9]|3[0-5])$
5-100       ^([5-9]|[1-8][0-9]|9[0-9]|100)$
12-1234     ^(1[2-9]|[2-9][0-9]|[1-8][0-9]{2}|9[0-8][0-9]|99[0-9]|1[01][0-9]{2}|12[0-2][0-9]|123[0-4])$
123-123     ^(123)$
256-321     ^(25[6-9]|2[6-9][0-9]|3[01][0-9]|32[01])$
256-257     ^(25[67])$
180-195     ^(18[0-9]|19[0-5])$

Answer 4

有理由必须是正则表达式吗？不能做这样的事情：

if ($number >= 256 && $number <= 321){
   // do something 
}

更新：

有一个简单但丑陋的方法可以使用 range 来做到这一点：

function getRangeRegex($from, $to)
{
    $range = implode('|', range($from, $to));

    // returns: 256|257|...|321
    return $range;
}

Answer 5

请小心，优秀的 @Bart Kiers 代码（以及 Travis J 的 JS 版本）在某些情况下会失败。例如：

12-1234    /^(?:1[2-9]|[2-9][0-9]|[1-9][0-9][0-9]|1[0-2][0-3][0-4])$/

不匹配“1229”、“1115”、“1[0-2][0-2][5-9]”

Answer 6

这实际上已经完成了。

请查看此网站。它包含一个 python 脚本的链接，该脚本会自动为您生成这些正则表达式。

Answer 7

continue

Answer 8

我已将 Bart Kiers 的答案转换为 C++。该函数将两个整数作为输入并生成数字范围的正则表达式。

#include <stdio.h>
#include <iostream>
#include <vector>
#include <string>

std::string regex_range(int from, int to);

int main(int argc, char **argv)
{
    std::string regex = regex_range(1,100);

    std::cout << regex << std::endl;

    return 0;
}

std::string regex_range(int from, int to) //Credit: Bart Kiers 2011
{
    if(from < 0 || to < 0)
    {
        std::cout << "Negative values not supported. Exiting." << std::endl;
        return 0;
    }

    if(from > to)
    {
        std::cout << "Invalid range, from > to. Exiting." << std::endl;
        return 0;
    }

    std::vector<int> ranges;
    ranges.push_back(from);
    int increment = 1;
    int next = from;
    bool higher = true;

    while(true)
    {

        next += increment;

        if(next + increment > to)
        {
            if(next <= to)
            {
                ranges.push_back(next);
            }
            increment /= 10;
            higher = false;
        }
        else if(next % (increment*10) == 0)
        {
            ranges.push_back(next);
            increment = higher ? increment*10 : increment/10;
        }

        if(!higher && (increment < 10))
        {
            break;
        }
    }

    ranges.push_back(to + 1);
    std::string regex("^(?:");

    for(int i = 0; i < ranges.size() - 1; i++)
    {
        int current_from = ranges.at(i);
        std::string str_from = std::to_string(current_from);
        int current_to = ranges.at(i + 1) - 1;
        std::string str_to = std::to_string(current_to);
        for(int j = 0; j < str_from.length(); j++)
        {
            if(str_from.at(j) == str_to.at(j))
            {
                std::string str_from_at_j(&str_from.at(j));
                regex.append(str_from_at_j);
            }
            else
            {
                std::string str_from_at_j(&str_from.at(j));
                std::string str_to_at_j(&str_to.at(j));

                regex.append("[");
                regex.append(str_from_at_j);
                regex.append("-");
                regex.append(str_to_at_j);
                regex.append("]");
            }
        }
        regex.append("|");
    }
    regex = regex.substr(0, regex.length() - 1);
    regex.append(")$");
    return regex;
}

Answer 9

因为我遇到同样的问题@EmilianoT 已经 报告，我尝试修复它，但最终我选择移植 rel="nofollow noreferrer">RegexNumericRangeGenerator （由 @EmilianoT 移植），尽管不在一个类中。我对这个 JS 端口不太满意，因为所有 toString() 和 parseInt() 方法仍然可以优化（它们可能在不必要的地方），但它适用于所有情况。

我改变的是参数。我用 parse(min, max, width = 0, prefix = '' 替换了 parse($min, $max, $MatchWholeWord = FALSE, $MatchWholeLine = FALSE, $MatchLeadingZero = FALSE) , suffix = '')，这给了它更多选项（有些人可能希望将正则表达式放入斜杠中，其他人希望匹配该行[前缀 = '^'; 后缀 = '$'] 等）。我还希望能够配置数字的宽度（width = 3 → 000、001、052 >、800、1000、...）。

我替换了之前的答案，因为它并不总是有效。如果有人想阅读它，他们可以在答案历史记录中看到它。

function parse(min, max, width = 0, prefix = '', suffix = '') {
  if (! Number.isInteger(min) || ! Number.isInteger(max) || min > max || min < 0 || max < 0) {
    return false
  }

  if (min == max) {
    return parseIntoPattern(min, prefix, suffix)
  }

  let x = parseStartRange(min, max)
  let s = []

  x.forEach(o => {
    s.push(parseEndRange(o[0], o[1]))
  })

  let n = reformatArray(s)
  let h = parseIntoRegex(n, width)

  return parseIntoPattern(h, prefix, suffix)
}

function parseIntoPattern(t, prefix = '', suffix = '') {
  let r = Array.isArray(t) ? t.join('|') : t
  return prefix + '(' + r + ')' + suffix
}

function parseIntoRegex(t, width = 0) {
  if (! Array.isArray(t)) {
    throw new Error('Argument needs to be an array!')
  }

  let r = []

  for (let i = 0; i < t.length; i++) {
    let e = t[i][0].split('')
    let n = t[i][1].split('')
    let s = ''
    let o = 0
    let h = ''

    for (let a = 0; a < e.length; a++) {
      if (e[a] === n[a]) {
        h += e[a]
      } else if (parseInt(e[a]) + 1 === parseInt(n[a])) {
        h += '[' + e[a] + n[a] + ']'
      } else {
        if (s === e[a] + n[a]) {
          o++
        }

        s = e[a] + n[a]

        if (a == e.length - 1) {
          h += o > 0 ? '{' + (o + 1) + '}' : '[' + e[a] + '-' + n[a] + ']'
        } else if (o === 0) {
          h += '[' + e[a] + '-' + n[a] + ']'
        }
      }
    }

    if (e.length < width) {
      h = '0'.repeat(width - e.length, '0') + h
    }

    r.push(h)
  }

  return r
}

function reformatArray(t) {
  let arrReturn = []

  for (let i = 0; i < t.length; i++) {
    let page = t[i].length / 2

    for (let a = 0; a < page; a++) {
      arrReturn.push(t[i].slice(2 * a))
    }
  }

  return arrReturn
}

function parseStartRange(t, r) {
  t = t.toString()
  r = r.toString()

  if (t.length === r.length) {
    return [[t, r]]
  }

  let breakOut = 10 ** t.length - 1
  return [[t, breakOut.toString()]].concat(parseStartRange(breakOut + 1, r))
}

function parseEndRange(t, r) {
  if (t.length == 1) {
    return [t, r]
  }

  if ('0'.repeat(t.length) === '0' + t.substr(1)) {
    if ('0'.repeat(r.length) == '9' + r.substr(1)) {
      return [t, r]
    }

    if (parseInt(t.toString().substr(0, 1)) < parseInt(r.toString().substr(0, 1))) {
      let e = parseInt(r.toString().substr(0, 1) + '0'.repeat(r.length - 1)) - 1
      return [t, strBreakPoint(e)].concat(parseEndRange(strBreakPoint(e + 1), r))
    }
  }

  if ('9'.repeat(r.length) === '9' + r.toString().substr(1) && parseInt(t.toString().substr(0, 1)) < parseInt(r.toString().substr(0, 1))) {
    let e = parseInt(parseInt(parseInt(t.toString().substr(0, 1)) + 1) + '0'.repeat(r.length - 1)) - 1
    return parseEndRange(t, strBreakPoint(e)).concat(strBreakPoint(e + 1), r)
  }

  if (parseInt(t.toString().substr(0, 1)) < parseInt(r.toString().substr(0, 1))) {
    let e = parseInt(parseInt(parseInt(t.toString().substr(0, 1)) + 1) + '0'.repeat(r.length - 1)) - 1
    return parseEndRange(t, strBreakPoint(e)).concat(parseEndRange(strBreakPoint(e + 1), r))
  }

  let a = parseInt(t.toString().substr(0, 1))
  let o = parseEndRange(t.toString().substr(1), r.toString().substr(1))
  let h = []

  for (let u = 0; u < o.length; u++) {
    h.push(a + o[u])
  }

  return h
}

function strBreakPoint(t) {
  return t.toString().padStart((parseInt(t) + 1).toString().length, '0')
}